hi all,
i am trying to see if there is a simplified expression for a
particular random variable using mathematica. i have two independent
binomial random variables, X and Y, and i'd like to compute the
expectation of X/(X+Y). more formally, X is distributed Binomial(p1,
n) and Y is distributed Binomial(p2, n). note that p1 and p2 can be
different, but the n's are the same. i would like to compute E(X/(X
+Y)). since the ratio is undefined when X+Y = 0, i would like to
assume that X+Y >= 1.
is there a way to compute this expectation and see if it has a
simplified analytic form subject to this constraint?
i defined my function to compute the probability of this random
variable as follows:
myvar[x_, y_] :=
Binomial[n, x]*p1^x*(1 - p1)^(n - x)*Binomial[n, y]*
p2^y*(1 - p2)^(n - y)*(x/(x + y))
now i just want to take the expectation of this. since the value of X
ranges from 0 to N and the value of Y ranges from 0 to N, it should be
possible to simply take a sum over X from 0 to N and sum of Y from 0
to Y to see what this expression evaluates to and whether it has a
simple analytic form.
the trick is to take the sum with the constraint of X + Y >= 1 -- how
can i express this in Mathematica?
i am also open to computing the expectation of X/(X+Y) by assuming
they are independent Poissons if that makes the math easier, i.e.
using poisson approximation to binomial.
thank you for your help.
> > where X, Y are independent but p1 and p2 could be different. I am
> > trying to compute the mean E(X/(X+Y)) but i am stuck. is there an easy
> > form for this mean? i looked it up in several probability books but
> > could not get it.
>
> > i am willing to make either a Poisson approximation or a Normal
> > approximation to thebinomialfor my application if i could compute
> > this mean. if i assume that X and Y are distributed Poisson, i know
> > that E(X+Y) has a nice form as a function of the rates of the two
> > poissons, but i still do not know how to get the distribution of X/(X
> > +Y) so that i can compute the expectation E(X/(X+Y)).
>